Taking real and imaginary parts after reciprocalpresenting a real number as real instead of imaginaryAssumptions about list elementsIntegrate yields complex value, while after variable transformation the result is real. Bug?Assumptions and Conditions in Integrate Function in MathmaticaAssumption that a parameter is realUnable to simplify the real part of a complex expressionHow to check the assumptions attached to a given symbol?How can I tell the kernel that a variable is Real?Assuming and Conjugate do not work as expectedSeries default assumptions?
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Taking real and imaginary parts after reciprocal
presenting a real number as real instead of imaginaryAssumptions about list elementsIntegrate yields complex value, while after variable transformation the result is real. Bug?Assumptions and Conditions in Integrate Function in MathmaticaAssumption that a parameter is realUnable to simplify the real part of a complex expressionHow to check the assumptions attached to a given symbol?How can I tell the kernel that a variable is Real?Assuming and Conjugate do not work as expectedSeries default assumptions?
$begingroup$
I noticed the following strange scenario. When I defined a variable to be real, Mathematica does not only recognize that it is real after taking an inverse. How can I resolve this so that it will recognize 1/x is still real?
In[29]:= $Assumptions = x ∈ Reals
Out[29]= x ∈ Reals
In[30]:= Im[x]
Out[30]= Im[x]
In[31]:= Refine[Im[x]]
Out[31]= 0
In[32]:= Refine[Re[x]]
Out[32]= x
In[33]:= Refine[Im[1/x]]
Out[33]= Im[1/x]
In[34]:= Refine[Re[1/x]]
Out[34]= Re[1/x]
complex assumptions
edited 7 hours ago
Michael E2
153k12208498
asked 9 hours ago
JamesJames
111
New contributor
James is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I noticed the following strange scenario. When I defined a variable to be real, Mathematica does not only recognize that it is real after taking an inverse. How can I resolve this so that it will recognize 1/x is still real?
In[29]:= $Assumptions = x ∈ Reals
Out[29]= x ∈ Reals
In[30]:= Im[x]
Out[30]= Im[x]
In[31]:= Refine[Im[x]]
Out[31]= 0
In[32]:= Refine[Re[x]]
Out[32]= x
In[33]:= Refine[Im[1/x]]
Out[33]= Im[1/x]
In[34]:= Refine[Re[1/x]]
Out[34]= Re[1/x]
complex assumptions
edited 7 hours ago
Michael E2
153k12208498
asked 9 hours ago
JamesJames
111
New contributor
James is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Another way to do it is useComplexExpandwithout any assumptions. For example:$Assumptions = True; ReIm[ 1 / x ] // ComplexExpand
$endgroup$
– LouisB
8 hours ago
1
$begingroup$
You can format inline code and code blocks by selecting the code and clicking thebutton above the edit window. The edit window help button?is useful for learning how to format your questions and answers. You may also find this meta Q&A helpful
$endgroup$
– Michael E2
7 hours ago
add a comment |
$begingroup$
I noticed the following strange scenario. When I defined a variable to be real, Mathematica does not only recognize that it is real after taking an inverse. How can I resolve this so that it will recognize 1/x is still real?
In[29]:= $Assumptions = x ∈ Reals
Out[29]= x ∈ Reals
In[30]:= Im[x]
Out[30]= Im[x]
In[31]:= Refine[Im[x]]
Out[31]= 0
In[32]:= Refine[Re[x]]
Out[32]= x
In[33]:= Refine[Im[1/x]]
Out[33]= Im[1/x]
In[34]:= Refine[Re[1/x]]
Out[34]= Re[1/x]
complex assumptions
edited 7 hours ago
Michael E2
153k12208498
asked 9 hours ago
JamesJames
111
New contributor
James is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I noticed the following strange scenario. When I defined a variable to be real, Mathematica does not only recognize that it is real after taking an inverse. How can I resolve this so that it will recognize 1/x is still real?
In[29]:= $Assumptions = x ∈ Reals
Out[29]= x ∈ Reals
In[30]:= Im[x]
Out[30]= Im[x]
In[31]:= Refine[Im[x]]
Out[31]= 0
In[32]:= Refine[Re[x]]
Out[32]= x
In[33]:= Refine[Im[1/x]]
Out[33]= Im[1/x]
In[34]:= Refine[Re[1/x]]
Out[34]= Re[1/x]
complex assumptions
complex assumptions
edited 7 hours ago
Michael E2
153k12208498
asked 9 hours ago
JamesJames
111
New contributor
James is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 7 hours ago
Michael E2
153k12208498
asked 9 hours ago
JamesJames
111
New contributor
James is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 7 hours ago
Michael E2
153k12208498
edited 7 hours ago
Michael E2
153k12208498
edited 7 hours ago
Michael E2
153k12208498
153k12208498
asked 9 hours ago
JamesJames
111
New contributor
James is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 9 hours ago
JamesJames
111
asked 9 hours ago
JamesJames
111
111
New contributor
James is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
James is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
Another way to do it is useComplexExpandwithout any assumptions. For example:$Assumptions = True; ReIm[ 1 / x ] // ComplexExpand
$endgroup$
– LouisB
8 hours ago
1
$begingroup$
You can format inline code and code blocks by selecting the code and clicking thebutton above the edit window. The edit window help button?is useful for learning how to format your questions and answers. You may also find this meta Q&A helpful
$endgroup$
– Michael E2
7 hours ago
add a comment |
1
$begingroup$
Another way to do it is useComplexExpandwithout any assumptions. For example:$Assumptions = True; ReIm[ 1 / x ] // ComplexExpand
$endgroup$
– LouisB
8 hours ago
1
$begingroup$
You can format inline code and code blocks by selecting the code and clicking thebutton above the edit window. The edit window help button?is useful for learning how to format your questions and answers. You may also find this meta Q&A helpful
$endgroup$
– Michael E2
7 hours ago
1
1
$begingroup$
Another way to do it is use
ComplexExpand without any assumptions. For example: $Assumptions = True; ReIm[ 1 / x ] // ComplexExpand$endgroup$
– LouisB
8 hours ago
$begingroup$
Another way to do it is use
ComplexExpand without any assumptions. For example: $Assumptions = True; ReIm[ 1 / x ] // ComplexExpand$endgroup$
– LouisB
8 hours ago
1
1
$begingroup$
You can format inline code and code blocks by selecting the code and clicking the
? is useful for learning how to format your questions and answers. You may also find this meta Q&A helpful$endgroup$
– Michael E2
7 hours ago
$begingroup$
You can format inline code and code blocks by selecting the code and clicking the
? is useful for learning how to format your questions and answers. You may also find this meta Q&A helpful$endgroup$
– Michael E2
7 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If x=0, then 1/0 is ComplexInfinity. If you add the assumption that x!=0, then you get what you are wanting:
Assuming[b [Element] Reals, b != 0, Refine[Im[1/b]]]
0
answered 8 hours ago
Robert JacobsonRobert Jacobson
834814
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
If x=0, then 1/0 is ComplexInfinity. If you add the assumption that x!=0, then you get what you are wanting:
Assuming[b [Element] Reals, b != 0, Refine[Im[1/b]]]
0
answered 8 hours ago
Robert JacobsonRobert Jacobson
834814
$endgroup$
add a comment |
$begingroup$
If x=0, then 1/0 is ComplexInfinity. If you add the assumption that x!=0, then you get what you are wanting:
Assuming[b [Element] Reals, b != 0, Refine[Im[1/b]]]
0
answered 8 hours ago
Robert JacobsonRobert Jacobson
834814
$endgroup$
add a comment |
$begingroup$
If x=0, then 1/0 is ComplexInfinity. If you add the assumption that x!=0, then you get what you are wanting:
Assuming[b [Element] Reals, b != 0, Refine[Im[1/b]]]
0
answered 8 hours ago
Robert JacobsonRobert Jacobson
834814
$endgroup$
If x=0, then 1/0 is ComplexInfinity. If you add the assumption that x!=0, then you get what you are wanting:
Assuming[b [Element] Reals, b != 0, Refine[Im[1/b]]]
0
answered 8 hours ago
Robert JacobsonRobert Jacobson
834814
answered 8 hours ago
Robert JacobsonRobert Jacobson
834814
answered 8 hours ago
Robert JacobsonRobert Jacobson
834814
answered 8 hours ago
Robert JacobsonRobert Jacobson
834814
834814
add a comment |
add a comment |
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1
$begingroup$
Another way to do it is use
ComplexExpandwithout any assumptions. For example:$Assumptions = True; ReIm[ 1 / x ] // ComplexExpand$endgroup$
– LouisB
8 hours ago
1
$begingroup$
You can format inline code and code blocks by selecting the code and clicking the
button above the edit window. The edit window help button?is useful for learning how to format your questions and answers. You may also find this meta Q&A helpful$endgroup$
– Michael E2
7 hours ago